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a) Ta có: \(\frac{x-2}{15}=\frac{9}{5}\Rightarrow5.\left(x-2\right)=9.15\)
\(\Rightarrow5x-10=135\)
\(\Rightarrow5x=145\)
\(\Rightarrow x=29\)
b) \(\frac{2-x}{16}=\frac{-4}{x-2}\Rightarrow\left(2-x\right).\left(x-2\right)=\left(-4\right).16\)
\(\Rightarrow4x-x^2-4=-64\)
\(\Rightarrow4x-x^2=-60\)
Lập bảng rồi tính ra
c) \(\frac{14}{x}=\frac{x-1}{4}\Rightarrow x.\left(x-1\right)=14.4\)
\(\Rightarrow x.\left(x-1\right)=56\)
Vì 56 = 8 x 7
\(\Rightarrow x=8\)
Câu c) còn có thêm \(x=-7\) nữa nha MMS_Hồ Khánh Châu vì \(x\inℤ\) mà :')
\(\frac{x}{-7}=\frac{5}{-35}\)
\(\frac{x.5}{-35}=\frac{5}{-35}\)
=> x . 5 = 5
x = 5 : 5
x = 1
\(\frac{x}{2}=\frac{3}{9}\)
\(\Leftrightarrow\frac{x}{2}=\frac{1}{3}\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\frac{2}{3}\)
b) \(\frac{x}{4}=\frac{-15}{12}=\frac{y+4}{16}\)(có lẽ đề như vậy)
\(\Rightarrow\frac{x}{4}=\frac{-15}{12}\)
\(\Rightarrow12x=-60\)
\(\Rightarrow x=-\frac{60}{12}=-5\)
\(\Rightarrow\frac{x}{4}=\frac{y+4}{16}\)
\(\Rightarrow16x=4y+16\)
\(\Rightarrow-80=4y+16\)
\(\Rightarrow4y=-96\)
\(\Rightarrow y=24\)
câu a hoàn toàn tương tự nên bạn làm nốt nhé!
a)Ta có:
\(\frac{y}{35}=-\frac{6}{14}\Rightarrow y=\frac{-6.35}{14}=-15\)
\(\frac{3}{x}=-\frac{15}{35}\Rightarrow x=\frac{3.35}{-15}=-7\)
Vậy: x=-15;y=-7
b)ta có:
\(\frac{x}{4}=-\frac{15}{12}\Rightarrow x=\frac{-15.4}{12}=-5\)
\(-\frac{15}{12}=\frac{y+4}{16}\Rightarrow y+4=\frac{-15.16}{12}=-20\Rightarrow y=-20-4=-24\)
Vậy x=-5; y=-24
\(a.\frac{x}{7}=\frac{x+16}{35}\)
\(\Rightarrow35x=7\left(x+16\right)\)
\(\Rightarrow35x=7x+112\)
\(\Rightarrow35x-7x=112\)
\(\Rightarrow28x=112\)
\(\Rightarrow x=112:28\)
\(\Rightarrow x=4\)
\(b.\frac{2-x}{16}=\frac{-4}{x-2}\)
\(\Rightarrow\frac{-\left(x-2\right)}{16}=\frac{-4}{x-2}\)
\(\Rightarrow\frac{x-2}{16}=\frac{4}{x-2}\)
\(\Rightarrow\left(x-2\right)^2=4.16\)
\(\Rightarrow\left(x-2\right)^2=64\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm8\right)^2\)
\(\Rightarrow x-2=8\) hoặc \(x-2=-8\)
+) Nếu \(x-2=8\)
\(\Rightarrow x=8+2\)
\(\Rightarrow x=10\)
+) Nếu \(x-2=-8\)
\(\Rightarrow x=-8+2\)
\(\Rightarrow x=-6\)
Vậy \(x=10;x=-6\)
Bài 1:
a) \(\frac{25}{4}+\frac{-5}{4}=\frac{25-5}{4}=\frac{20}{4}=5\)
b)\(\frac{-5}{9}+\left(\frac{-2}{7}\right)=\frac{-35}{63}+\left(\frac{-18}{63}\right)=\frac{-53}{63}\)
c) \(\frac{1}{4}+\frac{9}{11}+\frac{7}{4}+\left(\frac{-2}{11}\right)=\left(\frac{1}{4}+\frac{7}{4}\right)+\left(\frac{-2}{11}+\frac{9}{11}\right)=2+\frac{7}{11}=\frac{22+7}{11}=\frac{29}{11}\)
d) \(\frac{-5}{19}.\frac{8}{19}+\left(\frac{-14}{19}\right).\frac{11}{19}=\frac{-40}{361}-\frac{151}{361}=-\frac{191}{361}\)
Bài 2:
a) \(x+\frac{5}{9}=\frac{-8}{9}\) \(\Leftrightarrow x=\frac{-8}{9}-\frac{5}{9}\) \(\Leftrightarrow x=-\frac{13}{9}\)
b) \(\frac{-1}{8}-x=\frac{9}{20}\) \(\Leftrightarrow x=\frac{-1}{8}-\frac{9}{20}\) \(\Leftrightarrow x=\frac{-5}{40}-\frac{18}{40}\) \(\Leftrightarrow x=-\frac{23}{40}\)
c) (x + 5)3 - 12 = 15
\(\Leftrightarrow\)(x + 5)3 = 27
\(\Leftrightarrow\)x + 5 = 3
\(\Leftrightarrow\)x = -2
d) \(\left|x-3\right|-\frac{4}{15}=\frac{26}{15}\) \(\Leftrightarrow\left|x-3\right|=2\) \(\Leftrightarrow\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
a) \(\dfrac{x-2}{15}=\dfrac{9}{5}\)
\(\Leftrightarrow x-2=\dfrac{9\times15}{5}\)
\(\Leftrightarrow x-2=27\)
\(\Leftrightarrow x=29\)
Vậy ...............
b) \(\dfrac{2-x}{16}=\dfrac{-4}{x-2}\)
\(\Leftrightarrow\dfrac{2-x}{16}=\dfrac{4}{2-x}\)
\(\Leftrightarrow\left(2-x\right)^2=64\)
\(\Leftrightarrow\left(2-x\right)^2-64=0\)
\(\Leftrightarrow\left(2-x\right)^2-8^2=0\)
\(\Leftrightarrow\left(2-x-8\right)\left(2-x+8\right)=0\)
\(\Leftrightarrow\left(-x-6\right)\left(-x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x-6=0\\-x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=6\end{matrix}\right.\)
Vậy ..............
c) \(\dfrac{14}{x}=\dfrac{x-1}{4}\)
\(\Leftrightarrow x\left(x-1\right)=56\)
\(\Leftrightarrow x^2-x-56=0\)
\(\Leftrightarrow x^2+7x-8x-56=0\)
\(\Leftrightarrow x\left(x+7\right)-8\left(x+7\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=8\end{matrix}\right.\)
Vậy ....................
a) \(\dfrac{x-2}{15}=\dfrac{9}{5}\)
\(\Leftrightarrow\left(x-2\right).5=9.15\)
\(\Leftrightarrow5x-10=45\)
\(\Leftrightarrow5x=55\)
\(\Leftrightarrow x=11\) ( t/m x thuộc Z )
Vậy....